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A horizontal force, just sufficient to move a body of mass 4kg lying on a rough horizontal surface is applied on it. The coefficients of static and kinetic friction between the body and the surface are 0.8 and 0.6 respectively. If the force continues to act even after the block has started moving, the acceleration of the block in ms−2 is [g = 10 m s − 2 ]
- a)1/4
- b)1/2
- c)2
- d)4
Correct answer is option 'C'. Can you explain this answer?
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A horizontal force, just sufficient to move a body of mass 4kg lying o...
-2 will be 2 m/s^2.
Explanation:
The maximum static frictional force that can act on the block is given by:
fs = μsN
where μs is the coefficient of static friction and N is the normal force acting on the block. Since the block is at rest initially, the force applied must be equal to the maximum static frictional force:
F = fs = μsN
Substituting the values, we get:
F = 0.8 x 4 x 9.8 = 31.36 N
This is the maximum force that can be applied before the block starts moving. Once the block starts moving, the force of friction acting on it becomes kinetic friction and is given by:
fk = μkN
where μk is the coefficient of kinetic friction. The force applied is still greater than the force of kinetic friction, so the block continues to move with a constant velocity. The force required to maintain this velocity is equal to the force of kinetic friction:
F = fk = μkN
Substituting the values, we get:
F = 0.6 x 4 x 9.8 = 23.52 N
The net force acting on the block is given by:
Fnet = F - fk = 31.36 - 23.52 = 7.84 N
Using Newton's second law, we can find the acceleration of the block:
Fnet = ma
a = Fnet/m = 7.84/4 = 1.96 m/s^2
However, the question asks for the acceleration when the force continues to act even after the block has started moving. In this case, the force of kinetic friction remains constant, but the net force acting on the block is now:
Fnet = F - fk = 31.36 - 23.52 = 7.84 N
This net force is greater than the force required to maintain the constant velocity, so there is a net force acting on the block causing it to accelerate. Using Newton's second law, we can find the acceleration:
Fnet = ma
a = Fnet/m = 7.84/4 = 1.96 m/s^2
Therefore, the acceleration of the block when the force continues to act even after it has started moving is 1.96 m/s^2.
Explanation:
The maximum static frictional force that can act on the block is given by:
fs = μsN
where μs is the coefficient of static friction and N is the normal force acting on the block. Since the block is at rest initially, the force applied must be equal to the maximum static frictional force:
F = fs = μsN
Substituting the values, we get:
F = 0.8 x 4 x 9.8 = 31.36 N
This is the maximum force that can be applied before the block starts moving. Once the block starts moving, the force of friction acting on it becomes kinetic friction and is given by:
fk = μkN
where μk is the coefficient of kinetic friction. The force applied is still greater than the force of kinetic friction, so the block continues to move with a constant velocity. The force required to maintain this velocity is equal to the force of kinetic friction:
F = fk = μkN
Substituting the values, we get:
F = 0.6 x 4 x 9.8 = 23.52 N
The net force acting on the block is given by:
Fnet = F - fk = 31.36 - 23.52 = 7.84 N
Using Newton's second law, we can find the acceleration of the block:
Fnet = ma
a = Fnet/m = 7.84/4 = 1.96 m/s^2
However, the question asks for the acceleration when the force continues to act even after the block has started moving. In this case, the force of kinetic friction remains constant, but the net force acting on the block is now:
Fnet = F - fk = 31.36 - 23.52 = 7.84 N
This net force is greater than the force required to maintain the constant velocity, so there is a net force acting on the block causing it to accelerate. Using Newton's second law, we can find the acceleration:
Fnet = ma
a = Fnet/m = 7.84/4 = 1.96 m/s^2
Therefore, the acceleration of the block when the force continues to act even after it has started moving is 1.96 m/s^2.
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A horizontal force, just sufficient to move a body of mass 4kg lying on a rough horizontal surface is applied on it. The coefficients of static and kinetic friction between the body and the surface are 0.8 and 0.6 respectively. If the force continues to act even after the block has started moving, the acceleration of the block in ms−2 is [g = 10 m s − 2 ]a)1/4b)1/2c)2d)4Correct answer is option 'C'. Can you explain this answer?
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A horizontal force, just sufficient to move a body of mass 4kg lying on a rough horizontal surface is applied on it. The coefficients of static and kinetic friction between the body and the surface are 0.8 and 0.6 respectively. If the force continues to act even after the block has started moving, the acceleration of the block in ms−2 is [g = 10 m s − 2 ]a)1/4b)1/2c)2d)4Correct answer is option 'C'. Can you explain this answer? for Class 10 2024 is part of Class 10 preparation. The Question and answers have been prepared according to the Class 10 exam syllabus. Information about A horizontal force, just sufficient to move a body of mass 4kg lying on a rough horizontal surface is applied on it. The coefficients of static and kinetic friction between the body and the surface are 0.8 and 0.6 respectively. If the force continues to act even after the block has started moving, the acceleration of the block in ms−2 is [g = 10 m s − 2 ]a)1/4b)1/2c)2d)4Correct answer is option 'C'. Can you explain this answer? covers all topics & solutions for Class 10 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A horizontal force, just sufficient to move a body of mass 4kg lying on a rough horizontal surface is applied on it. The coefficients of static and kinetic friction between the body and the surface are 0.8 and 0.6 respectively. If the force continues to act even after the block has started moving, the acceleration of the block in ms−2 is [g = 10 m s − 2 ]a)1/4b)1/2c)2d)4Correct answer is option 'C'. Can you explain this answer?.
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